Question 1119477
{{{k(x-1)^2=5x-7}}}
Algebraic steps lead to
{{{kx^2-(2x+5)x+(k+7)=0}}}.



If r is one root, then if as given one root is double the other, other root is 2r.


SUM OF ROOTS
{{{r+2r=(2k+5)/k}}}
{{{3r=(2k+5)/k}}}


PRODUCT OF ROOTS
{{{r*2r=(k+7)/k}}}
{{{2r^2=(k+7)/k}}}


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If the sum of roots equation is used to solve for r in terms of k, and then r^2 calculated and substituted into the product-of-roots equation, ... the algebra steps will lead to when simplified, {{{k^2+23k-50=0}}}.


This factorizes to   {{{(k-2)(k+25)=0}}}.